Question

我通过onetoone关系加入了3个实体。我的目标是使用match.status != null的hibernate条件获取实体。如何告诉hibernate不要将algo实体加入到结果中，应该是（pick.algo = null）。

@Entity
public class Pick {
  @Id
  @GeneratedValue(strategy=GenerationType.AUTO)
  private int pid;

  @Column(columnDefinition="DATETIME")
  private Date insertTime;

  @Column(columnDefinition="DATETIME")
  private Date setupResTime;

  @OneToOne
  private DbMatch match;

  @OneToOne
  private Algo algo;

  @Transient
  private Integer algoID;
  ....

标准查询：

public List<Pick> getPicksHistory(){
    Criteria criteria = session.createCriteria(Pick.class);  
    criteria.add(Restrictions.isNotNull("match.status"));
    return criteria.list();

}

Answer 1

您可以添加别名并将条件应用于别名

Criteria criteria = session.createCriteria(Pick.class);  
criteria.createAlias("match", "match", JoinType.INNER_JOIN); //<---
criteria.add(Restrictions.isNotNull("match.status"));

要获取空值，您可以使用例如JoinType.LEFT_OUTER_JOIN

Answer 2

来自hibernate文档（http://docs.jboss.org/hibernate/orm/4.3/manual/en-US/html_single/#querycriteria-associations）：

Criteria criteria = session.createCriteria(Pick.class);
criteria.createAlias("match", "m");
criteria.add(Restrictions.isNotNull("m.status"));
criteria.setFetchMode("algo", FetchMode.LAZY);
criteria.list();

如何在hibernate标准上使用连接列？

2 个答案: